Hello Aspirants. Welcome to Online Quantitative Aptitude Section with explanation in **AffairsCloud.com**. Here we are creating question sample in **LCM & HCF**, which is common for all the IBPS, SBI exam and other competitive exams. We have includedÂ questions that are repeatedly asked in bank exams !!!

**What is the HCF of 3/5, 7/10, 2/15, 6/21?**

A) 5/123

B) 6/121

C) 2/105

D) 1/210

E) None of these**D) 1/210**

Explanation:

HCF will be HCF of numerators/LCM of denominators

So HCF = HCF of 3,7,2,6)/(LCM of 5,10, 15, 21)

= 1/210**The HCF and LCM of two numbers is 84 and 840 respectively. If the first number is 168, find the second one.**

A) 210

B) 420

C) 225

D) 326

E) None of these**B) 420**

Explanation:

Product of two numbers = HCF * LCM

So 2nd number = 84*840/168**Sum of 2 numbers is 128 and their HCF is 8. How many numbers of pairs of numbers will satisfy this condition?**

A) 2

B) 5

C) 3

D) 6

E) 4**E) 4**

Explanation:

Since HCF = 8 is highest common factor among those numbers

So, let first no = 8x, 2nd number = 8y

So 8x + 8y = 128

x + y = 16

the co-prime numbers which sum up 16 are (1,15), (3,13), (5,11), and (7,9)

*co-prime numbers are those which do no have any factor in common. These pairs are taken here because in HCF highest common factor is taken, so the remaining multiples x and y must have no common factors.

Since there are 4 pairs of co-prime numbers here, there will be 4 pairs of numbers satisfying given conditions. These are (8*1, 8*15), (8*3, 8*13) , (8*5, 8*11) , (8*7, 8*9)**Find the least number exactly divisible by 10, 15, 18 and 30.**

A) 85

B) 90

C) 88

D) 93

E) 105**B) 90**

Explanation:

It will be the LCM of these numbers.**Find the least number which when divided by15, 21, 24 and 32 leaves the same remainder 2 in each case.**

A) 3362

B) 3360

C) 3456

D) 3266

E) 3364**A) 3362**

Explanation:

It will be the LCM of these numbers + 2

So 3360 + 2**Find the least number which when divided by 10, 15, 18 and 30 leaves remainders 6, 11, 14 and 26 respectively.**

A) 72

B) 86

C) 85

D) 90

E) 94**B) 86**

Explanation:

Since 10-6 = 4, 15-11 = 4, 18-14 = 4, 30-26 = 4

So answer will be the LCM of these numbers â€“ 4

So 90 – 4**Find the smallest number of 4 digits which is exactly divisible by 12, 15, 21 and 30.**

A) 1580

B) 1420

C) 1260

D) 1056

E) None of these**C) 1260**

Explanation:

Least 4 digit number = 1000

LCM of (12, 15, 21, 30) = 420

On dividing 420 by 1000, leaves remainder = 160

So answer = 1000 + (420 â€“ 160)**Find the least number which when divided by 2, 5, 9 and 12 leaves a remainder 3 but leaves no remainder when same number is divided by 11.**

A) 281

B) 357

C) 360

D) 363

E) 224**D) 363**

Explanation:

LCM(2, 5, 9, 12) = 180

So the number will be somewhat = 180x + 3

This number leaves no remainder when divided by 11, so x = 2 to get (180x+3) fully divided by 11

So number = 180*2 + 3**HCF and LCM of two numbers is 5 and 275 respectively and the sum of these two numbers is 80. Find the sum of the reciprocals of these numbers.**

A) 15/242

B) 16/275

C) 32/221

D) 20/268

E) None of these**B) 16/275**

Explanation:

x + y = 80, xy = 5*275 = 1375

sum of reciprocals = 1/x + 1/y = (x+y)/xy = 80/1375**The HCF and LCM of two numbers is 70 and 1050 respectively. If the first number is 210, find the second one.**

A) 450

B) 350

C) 510

D) 260

E) None of these**B) 350**

Explanation:

Produvt of two numbers = HCF * LCM

So 2nd number = 70*1050/210

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